Area of Region Bounded by y = x^2 and y = 5x+6

[sapiental]

Consider the region bounded by y = x^2, y = 5x+6, and the negative x-axis

Compute the area of this region.

Im somewhat confused by what they mean by the negative x-axis?

The points of intersection between the two functions are [-1,1] and [6,36]

A = Integral -1 to 6 (5x-6-x^2)?

I'd appreciate if someone evaluated this problem for me step by step for I have spent several hours on it now without any clue.. Thank You!

 

[Office_Shredder]

It should be the parabola is on the top right of the region, the line is on the top left of the region, and the x-axis is on the bottom of the region

 

[HallsofIvy]

Consider the region bounded by y = x^2, y = 5x+6, and the negative x-axis

Compute the area of this region.

Im somewhat confused by what they mean by the negative x-axis?

The points of intersection between the two functions are [-1,1] and [6,36]

A = Integral -1 to 6 (5x-6-x^2)?

I'd appreciate if someone evaluated this problem for me step by step for I have spent several hours on it now without any clue.. Thank You!

Draw the graphs! The "negative x-axis" is the part of the x-axis left of (0,0), that corresponds to negative values of x. The integral you give is incorrect. It (with 5x+ 6, not 6x- 6) gives the area of the region between y= x² and y= 5x- 6. The area you want has upper boundary the straight line y= 5x- 6, from the point where it crosses the x-axis to (-1, 1) where it meets the parabola y= x², then that parabola to (0,0). The lower boundary is the x-axis, y= 0.

You will have to do it as to separate integrals.